good luck

Link: https://www.nowcoder.com/acm/contest/115/B
Source: Niuke.com

Time limit: C/C++ 1 second, other languages ​​2 seconds
Space limit: C/C++ 32768K, other languages ​​65536K
64bit IO Format: %lld

Topic description

Give a date of birth, for example: 1999-09-09,
ask: from the day of birth to today 2018-04-21 (including the date of birth and today), how many days, the year, month and day do not contain numbers 4?

Enter description:

Enter an integer T on the first line (representing the number of samples)
The next group T samples
One line per sample, containing a string "yyyy-mm-dd" (1990<=yyyy<=2018)
Questions to ensure the correctness of the test data

Output description:

Output the number of days required by the question

Example 1

enter

1
1999-09-09

output

5020

#include<iostream>
#include<cstdio>
using namespace std;
int a[2020][20][40];
void solve(){
        int cnt=0;
        for(int i=1990;i<=2018;i++){
            for(int j=1;j<=12;j++){
                if(j==2){
                    if(((i%4==0)&&(i%100!=0))||(i%400==0)){
                        for(int k=1;k<=29;k++){
                            if((i%10!=4)&&(j%10!=4)&&(k%10!=4))
                                cnt++;
                                
                                a[i][j][k]=cnt;
                        }
                    }
                    else{
                        for(int k=1;k<=28;k++){
                            if((i%10!=4)&&(j%10!=4)&&(k%10!=4))
                                cnt++;
                                a[i][j][k]=cnt;
                        }
                    }
                }
                else if(j==1||j==3||j==5||j==7||j==8||j==10||j==12){
                    for(int k=1;k<=31;k++){
                            if((i%10!=4)&&(j%10!=4)&&(k%10!=4))
                                cnt++;
                                a[i][j][k]=cnt;
                        }
                }
                else if(j==2||j==4||j==6||j==9||j==11){
                    for(int k=1;k<=30;k++){
                            if((i%10!=4)&&(j%10!=4)&&(k%10!=4))
                                cnt++;
                                a[i][j][k]=cnt;
                        }
                }
            }
        }
}
int main(){
    int n,x,y,z;
    cin>>n;
    solve();
    while(n--){
        scanf("%d-%d-%d",&x,&y,&z);
        int t=(a[2018][4][21]-a[x][y][z]);
        if((x%10!=4)&&(y%10!=4)&&(z%10!=4))
            t+=1;
        cout<<t<<endl;
    }
return 0;
}